The article \"An Investigation of the CaCO3-CaF2-K2SiO3-SiO2-Fe Flux System Usin

Question

The article "An Investigation of the CaCO3-CaF2-K2SiO3-SiO2-Fe Flux System Using the Submerged Arc Welding Process on HSLA-100 and A1SI-1081 Steels" (G. Fredrickson, M.S. Thesis, Colorado School of Mines, 1992) describes an experiment in which welding fluxes with differing chemical compositions were prepared. Several welds using each flux were made on A1S1-1018 steel base metal. The results of hardness measurements, on the Brinell, scale, of five welds using each of four fluxes are presented in Table below. Can we con elude that there are differences in the population means among the four flux types?

Explanation / Answer

this is a one-way anova model with factor as 'Flux type'.

We will assume equal variance of each sample and conduct the anova in R.

Code to make the dataset in R:

flux<-c(rep("A",5),rep("B",5),rep("C",5),rep("D",5))
flux
values<-c(250,264,256,260,239,263,254,267,265,267,
257,279,269,273,277,253,258,262,264,273)

fdata<-data.frame(flux,values)
fdata

Output:

> flux<-c(rep("A",5),rep("B",5),rep("C",5),rep("D",5))
> flux
[1] "A" "A" "A" "A" "A" "B" "B" "B" "B" "B" "C" "C" "C" "C" "C" "D" "D" "D" "D"
[20] "D"
> values<-c(250,264,256,260,239,263,254,267,265,267,
+ 257,279,269,273,277,253,258,262,264,273)
>
> fdata<-data.frame(flux,values)
> fdata
flux values
1 A 250
2 A 264
3 A 256
4 A 260
5 A 239
6 B 263
7 B 254
8 B 267
9 B 265
10 B 267
11 C 257
12 C 279
13 C 269
14 C 273
15 C 277
16 D 253
17 D 258
18 D 262
19 D 264
20 D 273
>

Now we will conduct one way anova with factor as the flux variable with the created dataset.

Code:

fdata$flux<-as.factor(fdata$flux)
anova(lm(values ~ flux, data=fdata))

Output:

> fdata$flux<-as.factor(fdata$flux)
> anova(lm(values ~ flux, data=fdata))
Analysis of Variance Table

Response: values
Df Sum Sq Mean Sq F value Pr(>F)
flux 3 743.4 247.800 3.8734 0.02944 *
Residuals 16 1023.6 63.975
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
>

As the p-value for flux is 0.029 which is less than 0.05, so at 5% significance level we reject null hypothesis of equality of population mean of the flux to conclude there is a significant difference in population mean among the four flux types.

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